A group consists of a set and a binary operation on that set that fulfills certain conditions. Groups are an example of example of algebraic structures, that all consist of one or more sets and operations on theses sets. The integers with the operations addition and multiplication are an example for another kind of algebraic structure, that consists of a set with two binary operation, that is a called a Ring. In this course we restrict our considerations to groups.
In the proceeding section we introduced all of the terminology needed to formally define a commutative group.
In the video in Figure 14.1 we motivate the definition of a group and give the definition. Following the video we present the formal definition of a group and,give examples.
Carefully read the definition.
Definition14.2.
A pair \((G,\bullet)\) consisting of a set \(G\) and a binary operation \(\bullet:G\times G\to G\) is a commutative group if the following properties hold:
Identity: There is an element \(e\in G\) such that for all \(a\in G\) we have \(a\bullet e=e\bullet a=a\text{.}\) The element \(e\) is called the identity of \((G,\bullet)\text{.}\)
Inverses: For each \(a\in G\) there is \(b\in G\) such that \(a\bullet b = b \bullet a= e\text{,}\) where \(e\) is the identity element in \(G\) with respect to \(\bullet\text{.}\) The element \(b\) is called an inverse of \(a\text{.}\)
Associativity: The operation \(\bullet\) is associative. So, \(a\bullet(b\bullet c) = (a\bullet b)\bullet c\) for all \(a\in G\text{,}\)\(b\in G\text{,}\) and \(c\in G\text{.}\)
Commutativity: The operation \(\bullet\) is commutative. So, \(a \bullet b=b \bullet a\) for all \(a\in G\) and \(b\in G\text{.}\)
Commutative groups are also called abelian groups after the Norwegian mathematician Niels Abel (1802 — 1829). A group that does not satisfy Item 4 is simply referred to as a group, or more specifically, a non-commutative group or non-abelian group. As we only consider commutative groups in this course, when we say group, we are referring to a commutative group. We call the operation \(\bullet\) of a group \((G,\bullet)\) the group operation of the group.
In Checkpoint 14.3 reproduce the definition of a group by filling in the blanks.
Checkpoint14.3.Group axioms.
A set S with a binary operation * on S is a commutative group if
\(\bullet\) there is
select
a complement
an element
an identity
an inverse
a set
an operation
with respect to * in S and
\(\bullet\) for each a in S there is
select
a complement
an element
an identity
an inverse
a set
an operation
with respect to * in S and
\(\bullet\) the operation * is
select
associative and commutative
associative and transitive
commutative and symmetric
.
In Theorem 13.19, we showed that a set with a binary operation has at most one identity element. So the identity element in a group is unique.
Similarly we can show that each element of a group has exactly one inverse with respect to the group operation \(\bullet\text{,}\) allowing us to speak of the inverse of an element. Recall that we denote the inverse of an element \(a\) with respect to the operation \(\bullet\) by \(a^{-1\bullet}\text{.}\)
Theorem14.4.
Let \((G,\bullet)\) be a group with identity element \(e \in G\text{.}\) Then, for each element \(a \in G\text{,}\) there is exactly one element \(b \in G\) such that \(a\bullet b=e\) and \(b\bullet a=e\text{,}\) implying that the inverse of each element \(a \in G\) is the unique element \(b=a^{-1\bullet}\text{.}\)
Proof.
Let \((G,\bullet)\) be a group with identity element \(e \in G\text{.}\) Suppose that \(b\in G\) and \(c\in G\) are both inverses of the element \(a\) in \((G,\bullet)\text{.}\) Then:
\begin{align*}
{2} b \amp = b \bullet e \qquad\qquad\amp \amp \text{ since \(e\) is the identity element of \((G,\bullet)\) }\\
\amp = b \bullet (a \bullet c) \qquad\qquad\amp \amp \text{ since \(a\) and \(c\) are inverses in \((G,\bullet)\) }\\
\amp = (b \bullet a) \bullet c \qquad\qquad\amp \amp \text{ since \((G,\bullet)\) is associative }\\
\amp = e \bullet c \qquad\qquad\amp \amp \text{ since \(a\) and \(b\) are inverses in \((G,\bullet)\) }\\
\amp = c \qquad\qquad\amp \amp \text{ since \(e\) is the identity element of \((G,\bullet)\) }
\end{align*}
Since \(b=c\text{,}\) there is exactly one inverse of \(a\) in \((G,\bullet)\text{,}\) and we write the inverse of \(a\) as \(a^{-1\bullet}\text{.}\)